"But this is for perfect blackbodies only, which have no theoretical existence."

Just for clarification, the Wien displacement law is not valid for emission spectrum "of blackbodies only". It is valid for any radiation in thermodynamic equilibrium with matter (equilibrium radiation), which is a more general concept that does not require presence of any black body. This equilibrium radiation has energy density spectrum given by the Planck function

${\rho}_{Planck}(\lambda ,T)=\frac{8\pi hc}{{\lambda}^{5}}\frac{1}{{e}^{\frac{hc}{\lambda kT}}-1}$

and can be produced and maintained by non-black bodies, for example by enclosure of any material brought to thermodynamic equilibrium at temperature $T$. Wien's displacement law says that the product of $T$ and ${\lambda}_{max}$ for which the energy density spectrum is maximum is the same for all temperatures.

"Does a similar formula exist for real bodies, which expresses λT in terms of its emissitivity ϵ?"

Let me rephrase your question : for emission intensity spectrum $I(\lambda ,T)$ of non-black body with known emissivity $\u03f5(\lambda ,T)$, is there a simple relation between the wavelength of maximum and temperature, similar to above Wien's displacement law ?

The answer is that in general, most probably no, because the emission radiation of real body has intensity spectrum proportional to

$I(\lambda ,T)\propto {\rho}_{Planck}(\lambda ,T)\u03f5(\lambda ,T),$

which means that the number, positions and height of peaks are modified by the emissivity $\u03f5(\lambda ,T)$. This depends on the body, its material and thickness and can be fairly general function of $\lambda $ and $T$. There may be some general restriction on this function due to physical laws, but Planck's formula does not imply any, apart from the condition that $0<\u03f5\le 1$ for all frequencies. One would have to study more detailed model of matter explaining value of $\u03f5(\lambda ,T)$ to find such relations.

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